Question

Problem 7-35 A satellite range prediction error has the standard normal distribution with mean 0 NM and standard deviation 1 NM. Find the following probabilities for the prediction error: (a) less than 1.17 NM Probability (b) between 1.04 NM and 1.99 NM Probability (c) greater than 0.93 NM Probability (d) between 0.24 NM and 0.73 NM Probability (e) less than or equal to 2.3 NM Probability (f) greater than 2.22 NM or less than .65 NM Probability (g) greater than or equal to 2.0 NM Probability\

Answer 1

Answer:

Kindly check explanation

Step-by-step explanation:

For a normal distribution with :

Mean(m) = 0

Standard deviation (s) = 1

Calculate :

(a) less than 1.17 NM Probability

Zscore = (x - m) / s = (1.17 - 0) / 1 = 1.17

P(z < 1.17) = 0.879

(b) between 1.04 NM and 1.99 NM Probability

Zscore = (x - m) / s = (1.04 - 0) / 1 = 1.04

P(z < 1.04) = 0.8508

Zscore = (x - m) / s = (1.99 - 0) / 1 = 1.99

P(z < 1.99) = 0.9767

(0.9767 - 0.8508) = 0.1259

(c) greater than 0.93 NM Probability

P(Z > 0.93) = 1 - p(Z < 0.93) = 1 - 0.8238 = 0.1762

(d) between 0.24 NM and 0.73 NM Probability

P(Z < 0.24) = 0.5948

P(Z <0.73) = 0.7673

0.7673 - 0.5948 = 0.1725

(e) less than or equal to 2.3 NM Probability

P(Z ≤ 2.3) = 0.9893

(f) greater than 2.22 NM or less than .65 NM

P(Z > 2.22) = 1 - p(Z < 2.22) = 1 - 0.9868 = 0.0132

P(Z < 0.65) = 0.7422

0.7422 - 0.0132 = 0.729

(g) greater than or equal to 2.0 NM Probability

P(Z ≥2) = 1 - p(Z ≤ 2)

1 - 0.97725 = 0.02275